ISRO HSFC Technician B Electrician Solved Question paper

Answer: "The maximum current the circuit breaker can interrupt "

The rupturing capacity (or breaking capacity) of a circuit breaker is the maximum short-circuit current that it can safely interrupt without being damaged or causing an unsafe condition.

• Expression: It is usually expressed in mega volt-amperes (MVA).
• Calculation: It is calculated as the product of the rated breaking current in kiloamperes (kA) and the rated voltage in kilovolts (kV).
• For example, a circuit breaker with a rated breaking current of 10 kA and a rated voltage of 20 kV would have a rupturing capacity of 200 MVA (10 kA * 20 kV). This means it can safely interrupt a fault current of 10 kA at a voltage of 20 kV.

Answer: "40000 "

Let’s break down the expression step by step:

Applying BODMAS Rule

Calculate 195 * 195:

195 * 195 = 38025

Calculate 2 * 195 * 5:

• 2 * 195 = 390
• 390 * 5 = 1950

Add the results from steps 1 and 2:

38025 + 1950 = 39975

Add 25:

39975 + 25 = 40000

Therefore, the value of the expression

195 x 195 + 2 x 195 x 5 + 25 is 40000.

Answer: "Diameter "

SWG, or Standard Wire Gauge, is used to measure the size (diameter) of conductors and the thickness of sheets. It is a system developed historically for wire diameter measurement, primarily used in Great Britain.

• Definition: SWG is a standardized system for measuring wire sizes, where a lower gauge number indicates a thicker wire.
• Applications: It is commonly used in various industries, including electrical wiring and manufacturing, to ensure compatibility and safety.
• Conversion: Although SWG remains relevant for specific applications, wire sizes are increasingly measured in metric units (millimeters) in modern contexts.

Answer: "All the above "

As we know that

• N ∝ Eb / Φ

This equation tells us that the speed of the motor (N) is directly proportional to the back EMF (Eb) and inversely proportional to the flux (Φ).

• N ∝ (V – IRa) / Φ

This equation is a modification of the first equation, where:

• V is the supply voltage
• I is the armature current
• Ra is the armature resistance

The speed of a DC motor can be controlled by varying several parameters. The correct options for controlling the speed of a DC motor are

1. Applied Voltage: The speed of a DC motor is directly proportional to the voltage applied across the armature. By increasing or decreasing this voltage, the speed of the motor can be controlled. By controlling the armature voltage below the rated value, the speed can be controlled only below the rated value.
   • Because the voltage can’t be above the rated value due to the insulation failure of winding.
   • This method is used for the separately excited motor but not suitable for the shunt motor because flux will change due to a change in voltage.
2. Flux per Pole: The speed is inversely proportional to the magnetic flux per pole. By varying the field current (which changes the flux), you can control the speed of the motor. Reducing the flux increases the speed, while increasing the flux decreases it.
3. Resistance of Armature Circuit: Introducing resistance in series with the armature reduces the current flowing through it, which in turn affects the back EMF and thus alters the speed of the motor. Increasing armature resistance will decrease speed, while decreasing resistance will increase speed.
   • Increasing the resistance of the armature circuit (Ra) will decrease the motor current (I) and consequently reduce the speed
   • This method gives only below base speeds.
   • In this method, the motor acts as constant torque and a variable power drive.

Answer: "Steel "

Specific resistance, also known as resistivity, is a fundamental property of materials that quantifies their ability to resist the flow of electric current. Here’s a more detailed overview:

• Specific Resistance (Resistivity): It is defined as the resistance of a uniform material with a unit length and unit cross-sectional area when a constant voltage is applied across it. The formula for resistivity (ρ) is given by:

$rho = R cdot frac{A}{L}$

Where:
R  = Resistance of the material (in ohms, Ω)
A = Cross-sectional area (in square meters, m²)
L = Length of the material (in meters, m)

Material	Specific Resistance (10-8 Ω⋅m)
Silver	1.59
Gold	2.44
Lead	22
Steel	46

 

Answer: "Yellow background with black border and symbol "

• Warning Signs: According to safety standards, warning signs are typically represented with a yellow background to indicate caution or a potential hazard. The symbols or text are usually in black to ensure high visibility and contrast.

• Standards Compliance: This color scheme is consistent with various safety regulations, including OSHA and ANSI guidelines, which specify that yellow is used for caution and warning signs.

Answer: "98.5 W "

Average Voltage to Peak Voltage

The relationship between average voltage ($V_{avg}$) and peak voltage ($V_p$) for a half-wave rectified sine wave is given by:

$V_{avg} = frac{V_p}{pi}$

Rearranging gives:

$V_p = V_{avg} times pi$

Given $V_{avg} = 40V$:

$V_p = 40 times pi approx 125.66V$

Calculating RMS Voltage

The RMS voltage ($V_{rms}$) for a half-wave rectified sine wave is calculated as:

$V_{rms} = frac{V_p}{2}$

Substituting the value of $V_p$:

$V_{rms} = frac{40pi}{2} = 20pi approx 62.83V$

Calculating Power Dissipated in the Resistor

The power P dissipated in a resistor is given by:

$P = frac{V_{rms}^2}{R}$

Substituting the value of V_rms and resistance R = 40Ω:

$P = frac{(20pi)^2}{40}$

Simplifying further:

$P = frac{400pi^2}{40} = 10pi^2$

Decimal Approximation of Power

Using the approximation π²≈ 9.8696 we can calculate:

$P approx 10 times 9.8696 approx 98.696W$

Therefore, the power dissipated in the resistor is approximately 98.69 Watts.

Answer: "Power factor of the motor "

A synchronous motor’s excitation level significantly influences its power factor.

Over-excitation:

• Leading Power Factor: When the motor is over-excited, it draws a leading current from the supply.
• Capacitive Effect: This leading current behaves similar to a capacitor, improving the power factor of the system.

Under-excitation:

• Lagging Power Factor: Conversely, under-excitation leads to a lagging power factor.
• Inductive Effect: The motor behaves like an inductive load, worsening the power factor.

Therefore, by adjusting the excitation level, we can control the power factor of a synchronous motor.

Answer: "m.R = n. r "

• There are m rows of cells, with each row containing nn cells connected in series.
• The internal resistance of each cell is r.
• The total internal resistance for one row of n cells in series is:

R_internal Resistance = nr.

• Since there are m such rows connected in parallel, the equivalent internal resistance Req​ of the entire configuration is given by:

Req = R_internal Resistance/m  = nr / m

Maximum Power Transfer Condition

According to the Maximum Power Transfer Theorem, maximum power is transferred to the load when the load resistance R is equal to the equivalent internal resistance of the source. Therefore, for maximum power transfer:

R = Req

Substituting for Req

R=n⋅r/m

Rearranging for Conditions

mR=n⋅r

Conclusion

Thus, the condition for maximum power transfer in this configuration is mR=n⋅r

 

Answer: "MHD generators "

An MHD generator (Magnetohydrodynamic generator) is a device that directly converts heat into electrical energy.

• Working Principle: MHD generators work on the principle of Faraday’s law of electromagnetic induction. This law states that when a conductor moves through a magnetic field, an electromotive force (EMF), or voltage, is induced in the conductor. In an MHD generator, the moving conductor is replaced by a high-temperature plasma, which is an electrically conductive gas.

• Components: The main components of an MHD generator are:

  • Plasma: A hot, ionized gas that acts as the moving conductor. It’s created by heating a gas (like argon or combustion products) to very high temperatures.
  • Magnetic Field: A strong magnetic field is applied perpendicular to the flow of the plasma.
  • Electrodes: Two electrodes are placed on opposite sides of the plasma flow, perpendicular to both the magnetic field and the plasma flow. These electrodes collect the induced electric current.